Methanol (CH3OH) is used as fuel in race cars and is a potential replacement for gasoline. It is synthesized by a combination of hydrogen and carbon monoxide. Suppose 65.8 g of carbon monoxide reacts with 8.6 g of hydrogen.
What is the limiting reagent?
1 Answer | Add Yours
First, let's look at the balanced chemical equation for the synthesis of methanol from carbon monoxide (CO) and hydrogen gas (H2).
CO + 2H2 --> CH3OH
Notice that we need two molecules of hydrogen gas for every molecule of CO to make a molecule of methanol. Now let's take the mass of each reagent and convert them to moles. We do this by dividing by the molecular weight for each compound.
65.8 g CO * (mole/44.01 g) = 1.50 moles CO
8.6 g H2 * (mole/2.02 g) = 4.26 moles H2
From the above equation we know that 2 moles of H2 are required for every mole of CO. Since we have 1.5 moles of CO, that would require 1.50*2=3 moles of H2 to completely consume it. We have 4.26 moles of H2 in actuality, so we will consume all of the CO present and still have 4.26-3=1.26 moles of H2 remaining. Since all of the CO will be consumed and H2 will remain, that makes CO the limiting reagent.
Can you please explain how you got the molecular weight of CO? From my calculations, I came to the answer of 28.01. That would be Carbon: 12.01 and Oxygen: 16. Is there something I'm missing or don't understand? Thank you!
We’ve answered 319,639 questions. We can answer yours, too.Ask a question